Question

Provide a sketch and state your assumptions as you determine the shape factor F12 for the following: Z-0.4 m 1 Zn = 0.2 m KY-0.3 mm X-0.5 m

Answer 1

Assumptions for determining shape factor-

1. The shape factor is purely a function of geometrical parameters.

2. When two bodies radiating energy with each other only, the shape factor relation is expressed as A₁ F_1-2 = A₂ F_2-1   

3. The shape factor of convex surface or flat surface with the other surface enclosing the first is always unity. This is because all the radiation coming out from the convex surface is intercepted by the enclosing surface but not vice versa.

4. A concave surface has a shape factor with itself because the radiation energy coming out from one part of the surface is intercepted by another part of the same surface. The shape factor of a surface with respect to itself is denoted by F_1-1.

                              F_1-1 = 0   for convex and flat surface.

If a surface of area A_l is completely enclosed by a second surface of area A₂ and if A_l does not see itself (F_1-1 = 0) then F_1-2 = l  

Then substituting the value of F_1-2 = l   in equation (1) ,

F 2-1 A A 2

Radiant exchange between coaxial cylinders, bodies placed in enclosures (sphere is kept in a room or box) are examples of this situation.

5. If n surfaces are taking part in radiation heat transfer then the energy radiated by one is always intercepted by the remaining (n-l) surfaces and by the surface itself also

               F_1-1 + F_1-2 + F_1-3 ……………F_1-n = 1

               F_2-1 + F_2-2 + F_2-3 ……………F_2-n = 1                                                                    (2)

               :          :                                     :

               :          :                                     :

               F_n-1 + F_n-2 + F_n-3 ……………F_n-n = 1

In addition to the above equations, the reciprocal relation between any two surfaces also holds good

                          F_(1-2) A₁ = F_(2-1) A₂                   or            F_(1-3) A₁ = F_(3-1) A₃     and so on.  

Now for solution

0.5 YX - 0.02 2 Z 1 0.05 Y x 0.4 10.1 10.2 0.31 10.4 0.6 0.2 1.0 15 2.00 10.1 4 10 2.01 0 0.1 0.2 0.4 N 4 0.6 0.81 ZX 6 8 10

Figure Configuration Factor for two perpendicular plates

Abso.am Zo=0:40 Y=030 Consider posts as shown in Fig. I have given names ③ & 4 apart from Azfag - A2 F 2 -3 + A2 F21 : 2-1 F2-4 - 12-3 Now see the o choot 724- 0.8 Y 06 a 0.5 from the chast F2-4 = 025 0.25 2 0,2 For 123 0.5 Y 0.3 os f2-3 = 0.16

fan = F2-4 - F2-3 0.25 -0.16 f2-1 : 0.09 Now , (F.. 2) (A) = (f2.,) A₂ (F1-2] (0.2x0.5) - (0.09) (0:340-5) Fi-2 o.ogxo.3 xos 0.2 xos Finzi 0135

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